Magnetic Moments: Aligning 90% with B at 0.2T

[t_n_p]

Homework Statement

A system of spin 1/2 magnet moments with μ = 1*10^-23 Am² are in a magnetic field of 0.2T. At what temperature will 90% of the magnets be aligned parallel to B?

The Attempt at a Solution

I think I have found the relevant equation:

U=-μB(N↑ - N↓)

where U is the magnetic potential energy, μ is the magnetic moment, B is the magnetic field, N↑ is the number of magnets oritentated up and N↓ is the number of magnets orientated down. My questions now are, how do I know which direction is parallel to B and if this equation is relevant, how do I then convert U to temperature?

 

[ravx]

From the way the question is phrased, I would believe the B-field is oriented in the "up" direction; this is because in general the energy of a magnetic dipole in a magnetic field is:

$$-\mu \cdot B$$

so the dipole will want to align with the field to get to the lowest energy. In this case, lower energy is achieved if B is pointing "up" rather than "down".

This much I know, the rest I'm not 100% sure, but I think the following would work:

Suppose we have N such spins, then we have $$N = N\uparrow - N\downarrow$$
Let $$2 S = N\uparrow - N\downarrow$$ be the spin difference. With this we can solve for
$$N\uparrow = S + \frac{N}{2}$$

and get

$$\frac{N\uparrow}{N} = \frac{S}{N} + \frac{1}{2}$$

Taking the average:


$$\frac{\left\langle N\uparrow \right\rangle}{N} = \frac{\left\langle S \right\rangle}{N} + \frac{1}{2}$$

We would like this ratio to be 90%.


For one single magnet, the partition function is given by

$$Z_{1} = 2 cosh (\beta\mu B)$$.

where $$\beta = \frac{1}{k_b T}$$

For N such magnets, our partition function will be

$$Z = \left(Z_{1}\right)^{N}$$

The average energy is given by:

$$\left\langle E \right\rangle = -\frac{1}{Z}\frac{\partial Z}{\partial\beta}$$

But we also know that $$\left\langle E \right\rangle = - \mu B \left\langle S \right\rangle$$

so now we can solve for the temperature "beta" at which our desired ratio is reached.